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BOWEN UNIVERSITY,IWO, OSUN STATE, NIGERIA.
STATISTICS PROGRAMME
TRANSPORTATION PROBLEM(PART B)
STA 322 (OPERATIONS RESEARCH) LECTURE NOTEOMOYAJOWO A.C
LEARNING OUTCOMES
At the end of the class, the students should be able to:
i. Solve maximization type of transportation problem
ii. Understand the concept of test for optimality in Transportation problem.
iii. Conveniently test for optimality in Transportation Problem.
Recap of Initial basic feasible solutionNorth west corner method Least cost method Vogel Approximation method
1. The allocation is madefrom the left hand side topcorner irrespective of the costof the cell.
The allocations are madedepending on the cost of thecell. Lowest cost is firstselected and then next highestetc.
The allocations are madedepending on the opportunitycost of the cell.
2. As no consideration isgiven to the cost of the cell,naturally the totaltransportation cost will behigher than the othermethods
As the cost of the cell isconsidered while makingallocations, the total cost oftransportation will becomparatively less.
As the allocations are madedepending on the opportunitycost of the cell, the basic feasiblesolution obtained will be verynearer to optimal solution.
Recap of Initial basic feasible solution
North west corner method Least cost method Vogel Approximation method
3. It takes less time. Thismethod is suitable to getbasic feasible solutionquickly.
The basic feasible solution, weget will be very nearer tooptimal solution. It takes moretime than northwest coronermethod.
It takes more time for gettingbasic Feasible solution. But thesolution we get will be verynearer to Optimal solution.
4. When basic feasiblesolution alone is asked, it isbetter to go for northwestcorner method.
When optimal solution isasked, better to go forinspection method for basicfeasible solution and MODI foroptimal solution.
VAM and MODI is the bestoption to get optimal solution.
Example on Maximization in Transportation Problem.
• A fertilizer company has three factories A,B,and C with annual capacities of 200, 500 and 300 thousand tonnes of urea. The product is distributed from four warehouses D,E, F, and G with annual offtake of 180, 320, 100 and tonnes of urea. Profit (in naira per thousand tonnes) is shown in table 1 . Find the optimum transportation schedule to maximize profit.
Example on Maximization in Transportation Problem.Table 1
Example on Maximization in Transportation Problem Cont’d
Step 1: Select the highest cost in the table (In this case the highest cost is 25).
Step 2: Subtract the other cost from the selected highest cost including itself (as seen in table 2).
Step 3: Solve the problem using the method of solving minimization problem(by finding the initial basic feasible solution and the optimal solution).
Example on Maximization in Transportation Problem Cont’d
TABLE 2 Destination
Sources D E F G Supply
A 13 17 19 0 200
B 17 18 15 7 500
c 11 22 14 5 300
Demand 180 320 100 400
STEP 2: MODI or Stepping Stone Method
Steps in MODIThe Modified Distribution Method, also known as MODI method or u-v method, which provides a minimum cost solution (optimal solution) to the transportation problem. The following are the steps involved in this method.
.
i and ii
Steps in MODI
Step 3: Any basic feasible solution has � + �-1 ��� ≥ 0. Thus, there will be � + � −
1 equation to
determine � + � dual variables. One of the dual variables can be chosen arbitrarily. It is also to be noted that as the primal constraints are equations, the dual variables are unrestricted in sign.
Step 4: If ��� = 0 the dual variables calculated in Step 3 are compared with the ���
values of this allocation as ��� −�� −��. If all ��� −�� −�� ≥ 0, then by the theorem
of complementary slackness it can be shown that the corresponding solution of the transportation problem is optimum. If one or more ��� −�� −�� < 0, we select the
cell with the least value of ��� −�� −�� and allocate as much as possible subject to
the row and column constraints. The allocations of the number of adjacent cell are adjusted so that a basic variable becomes non-basic.
Step 5: A fresh set of dual variables are calculated and repeat the entire procedure f
if it is not optimal, then steps 1-4 are repeated, till an optimal solution is obtained.
Steps in Stepping Stone methodThe Stepping stone method is used for finding the opportunity costs of empty cells. Every empty cell is to be evaluated for its opportunity cost. To do this the methodology is:
1. Put a small ‘+’ mark in the empty cell.
2. Starting from that cell draw a loop moving horizontally and vertically from loaded cell to loaded cell. Remember, there should not be any diagonal movement. We have to take turn only at loaded cells and move to vertically downward or upward or horizontally to reach another loaded cell. In between, if we have a loaded cell, where we cannot take a turn, ignore that and proceed to next loaded cell in that row or column.
3. After completing the loop, mark minus (–) and plus (+) signs alternatively.
4. Identify the lowest load in the cells marked with negative sign.
5. This number is to be added to the cells where plus sign is marked and subtract from the load of the cell where negative sign is marked.
Steps in Stepping Stone method Cont’d
6. Do not alter the loaded cells, which are not in the loop.
7. The process of adding and subtracting at each turn or corner is necessary to see that rim
requirements are satisfied.
8. Construct a table of empty cells and work out the cost change for a shift of load from loaded
cell to loaded cell.
9. If the cost change is positive, it means that if we include the evaluated cell in the programme,
the cost will increase. If the cost change is negative, the total cost will decrease, by including
the evaluated cell in the programme.
Steps in Stepping Stone method Cont’d
10. The negative of cost change is the opportunity cost. Hence, in the optimal solution of transportation problem empty cells should not have positive opportunity cost.
11. Once all the empty cells have negative opportunity cost, the solution is said to be optimal.
One of the drawbacks of stepping stone method is that we have to write a loop for every empty cell.
Hence it is tedious and time consuming. Therefore, for optimality test we use MODI method rather than the stepping stone method.
EXAMPLE ON TEST FOR OPTIMALITY
Consider the transportation problem with three sources and four destinations in table 3.
TABLE 3
Sources Destination Supply
1 2 3 4
1 2 3 11 7 6
2 1 0 6 1 1
3 5 8 15 9 10
Demand 7 5 3 2
EXAMPLE ON TEST FOR OPTIMALITY Cont’dUsing the Vogel Approximation method to obtain the Initial Basic Feasible Solution, we have table 4 :
TABLE 4
Sources Destination Supply
1 2 3 4
1 2 3 11 7 6
2 1 0 6 1 1 1
3 5 8 15 9 10
Demand 7 5 3 2
1 5
1
6 3 1
TEST FOR OPTIMALITY Cont’d
Here, � + � − 1 = 6 and the number of allocation is 6, therefore we can go ahead to use MODI to test for optimality. Then the opportunity cost for the empty cell is calculated as found in table 5:
TABLE 5BASIC VARIABLES Set �� = �, ��+�� = ��� NON- BASIC VARIABLES ��+ �� − ���
���: �� + �� = ��� = � + �� = �; �� = � ���: �� + �� = ��� = � + 12 − 11 = �
���: �� + �� = ��� = � + �� = �; �� = � ���: �� + �� = ��� = � + 6 − 7 = −�
���: �� + �� = ��� = �� + 6 = �; �� = � ���: �� + �� = ��� = � + 2 − 1 = �
���: �� + �� = ��� = �� + 2 = �; �� = � ���: �� + �� = ��� = � + 3 − 0 = �
���: �� + �� = ��� = � + �� = ��; �� = �� ���: �� + �� = ��� = � + �� − � = �
���: �� + �� = ��� = � + �� = �; �� = � ���: �� + �� = ��� = � + 3 − 8 = −�
Choose the most positive to enter the Basic Variable. Therefore, ��� enters
EXAMPLE ON TEST FOR OPTIMALITY Cont’d
• Now, to determine the leaving variable, a closed loop is made on the tableau with respect to the entering variable. The variable that gets to zero first, becomes the leaving variable( having the completeness in the supply and demand in mind).
TABLE 6
sources Destination Supply
1 2 3 4
1 2 3 11 7 6
2 1 0 6 1 1 1
3 5 8 15 9 10
Demand 7 5 3 2
1 5
6 3 1
1
3-1=2
1
1+1=2
1-1=0
EXAMPLE ON TEST FOR OPTIMALITY Cont’d
• We repeat the opportunity cost table to test for optimality and if the problem is not optimal, we choose the entering variable again.
TABLE 7
BASIC VARIABLES Set �� = �, ��+ �� = ��� NON BASIC VARIABLES ��+ �� − ���
���: �� + �� = ��� = � + �� = �; �� = � ���: �� + �� = ��� = � + 12 − 11 = �
���: �� + �� = ��� = � + �� = �; �� = � ���: �� + �� = ��� = � + 6 − 7 = −�
���: �� + �� = ��� = �� + 12 = �; �� = −� ���: �� + �� = ��� = −� + 2 − 1 = −�
���: �� + �� = ��� = �� + 2 = �; �� = � ���: �� + �� = ��� = −� + 3 − 0 = −�
���: �� + �� = ��� = � + �� = ��; �� = �� ���: �� + �� = ��� = −� + � − � = −�
���: �� + �� = ��� = � + �� = �; �� = � ���: �� + �� = ��� = � + 3 − 8 = −�
Choose the most positive to enter the Basic Variable. Therefore, ��� enters
EXAMPLE ON TEST FOR OPTIMALITY Cont’d
• The previous step to determine the leaving variable is done again by making a closed loop on the tableau with respect to the entering variable. The variable that gets to zero first becomes the leaving variable( having the completeness in the supply and demand in mind).
TABLE 8
sources Destination Supply
1 2 3 4
1 2 311 7
6
21 0 6 1
1
35 8 15 9
10
Demand 7 5 3 2
1 5
1
6 2 2
1
2-1=16+1=7
1-1=0
EXAMPLE ON TEST FOR OPTIMALITY Cont’d
• We repeat the opportunity cost table to test for optimality and if the problem is not optimal, we choose the entering variable again.
TABLE 9BASIC VARIABLES Set �� = �, ��+ �� = ��� NON BASIC VARIABLES ��+ �� − ���
���: �� + �� = ��� = � + �� = �; �� = � ���: �� + �� = ��� = � + 1 − 11 = −��
���: �� + �� = ��� = � + �� = ��; �� = �� ���: �� + �� = ��� = � + 5 − 7 = −�
���: �� + �� = ��� = �� + 1� = �; �� = −� ���: �� + �� = ��� = −� + 1 − 1 = −�
���: �� + �� = ��� = � + �� = �; �� = � ���: �� + �� = ��� = −� + 3 − 0 = −�
���: �� + �� = ��� = �� +�� = ��; �� = � ���: �� + �� = ��� = −� + � − � = −�
���: �� + �� = ��� = � + �� = �; �� = � ���: �� + �� = ��� = � + 3 − 8 = −�
Since all the variables are negative, optimality condition has been satisfied
EXAMPLE ON TEST FOR OPTIMALITY Cont’d• Therefore, the total minimum cost for transporting the goods from the
sources to the destinations are given below:
TABLE 10Interpretation Total cost
5 goods will be transported from source 1 to warehouse 1 at the rate of ₦3 per good
������ = � ∗� = 15
1 good will be transported from source 1 to warehouse 3 at the rate of ₦11per good
������ = � ∗�� = ��
1 good will be transported from source 2 to warehouse 3 at the rate of ₦6 per good
������ = � ∗� = �
7 goods will be transported from source 3 to warehouse 1 at the rate of ₦5 per good
������ = � ∗� = ��
1 good will be transported from source 3 to warehouse 3 at the rate of ₦15 per good
������ = � ∗�� = ��
2 goods will be transported from source 3 to warehouse 4 at the rate of ₦9 per good
������ = � ∗� = ��
Total ₦100
References
• Operations Research by Prem Kumar Gupta, D.S Hira .
• Operations Research by P.Rama Murthy - 2nd Edition.(b-ok.org)
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